5.6 Unforced Mechanical Vibrations
215
5.6 Unforced Mechanical Vibrations
The study of vibrating mechanical systems begins here with examples
for unforced systems with one degree of freedom. The main example is
a mass on a spring. The undamped, unforced cases are considered in a
number of physical examples, which include the following: simple pendulum, compound pendulum, swinging rod, torsional pendulum, shockless
auto, sliding wheel, rolling wheel.
Simple harmonic motion
Consider the spring-mass system of Figure 2, where x measures the
signed distance from the equilibrium position of the mass. The spring is
assumed to exert a force under both compression and elongation. Such
springs are commonly used in automotive suspension systems, notably
coil springs and leaf springs. In the case of coil springs, it is assumed
that there is space between the coils, which allows the spring to exert
bidirectional forces.
k
x<0
x=0
x>0
m
Figure 2. An undamped spring-mass
system, showing compression,
equilibrium and elongation of the
spring with corresponding positions of
the mass m.
Hooke’s law. The basic physical law to be applied is:
The linear restoring force F exerted by a spring is proportional to the signed elongation X, briefly, F = −kX.
The number k is called Hooke’s constant for the spring. In the model
of Figure 2, X = x(t) and k > 0. The minus sign accounts for the action
of the force: the spring tries to restore the mass to the equilibrium state,
so the vector force is directed toward the equilibrium position x = 0.
Newton’s second law. Specialized to the model in Figure 2, Newton’s
second law says:
The force F exerted by a mass m attached to a spring is
F = ma where a = d2 x/dt2 is the acceleration of the mass.
The weight W = mg is defined in terms of the gravitational constant g = 32 ft/s2 , 9.8 m/s2 or 980 cm/s2 where the mass m is given
respectively in slugs, kilograms or grams. The weight is the force due
to gravity and it has the appropriate units for a force: pounds in the
case of the fps system of units.
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Method of force competition. Hooke’s law F = −kx(t) and Newton’s second law F = mx′′ (t) give two independent equations for the force
acting on the system. Equating competing forces implies that the signed
displacement x(t) satisfies the free vibration equation
mx′′ (t) + kx(t) = 0.
It is also called the harmonic oscillator, especially in its equivalent
form
k
x′′ (t) + ω 2 x(t) = 0, ω 2 = .
m
In this context, ω is the natural frequency of the free vibration. The
harmonic oscillator is said to describe a simple harmonic motion x(t),
known by the recipe to have the form x(t) = c1 cos ωt + c2 sin ωt.
Phase-amplitude conversion. Given a simple harmonic motion
x(t) = c1 cos ωt + c2 sin ωt, as in Figure 3, define amplitude A and
phase angle α by the formulas
A=
q
c21 + c22 ,
c1 = A cos α
and c2 = A sin α.
Then the simple harmonic motion has the phase-amplitude form
(1)
x(t) = A cos(ωt − α).
To directly obtain (1) from trigonometry, use the trigonometric identity
cos(a − b) = cos a cos b + sin a sin b with a = ωt and b = α. It is known
from trigonometry that x(t) has period 2π/ω and phase shift α/ω.
A full period is called a cycle and a half-period a semicycle. The
frequency ω/(2π) is the number of complete cycles per second, or the
reciprocal of the period.
A
α
ω
−A
2π
ω
Figure 3. Simple harmonic
oscillation x(t) = A cos(ωt − α),
showing the period 2π/ω, the phase
shift α/ω and the amplitude A.
The phase shift is the amount of horizontal translation required to shift
the cosine curve cos(ωt − α) so that its graph is atop cos(ωt). To find
the phase shift from x(t), set the argument of the cosine term to zero,
then solve for t.
To solve for α ≥ 0 in the equations c1 = A cos α, c2 = A sin α, first
compute numerically by calculator the radian angle φ = arctan(c2 /c1 ),
which is in the range −π/2 to π/2. Quadrantial angle rules must be
applied when c1 = 0, because calculators return an error code for division
5.6 Unforced Mechanical Vibrations
217
by zero. A common error is to set α equal to φ. Not just the violation
of α ≥ 0 results — the error is a fundamental one, due to trigonmetric
intricacies, causing us to consider the equations c1 = A cos α, c2 = A sin α
in order to construct the answer for α:
α=


φ







(c1 , c2 )
φ + π (c1 , c2 )
φ + π (c1 , c2 )
φ + 2π (c1 , c2 )
in
in
in
in
quadrant
quadrant
quadrant
quadrant
I,
II,
III,
IV .
Visualization of harmonic motion. A simple harmonic motion
can be obtained graphically by means of the experiment shown in Figure
4, in which an undamped spring-mass system has an attached pen that
writes on a moving paper chart. The chart produces the simple harmonic
motion x(t) = c1 cos ωt + c2 sin ωt or equivalently x(t) = A cos(ωt − φ).
pen
paper
mass
motion
Figure 4. A moving paper chart
records the vertical motion of a mass
on a spring by means of an attached
pen.
Applications
We will consider various incarnations of the pendulum. The illustrations
end with applications to auto suspension systems and rolling wheels.
The simple pendulum. A pendulum is constructed from a thin
massless wire or rod of length L and a body of mass m, as in Figure 5.
Along the circular arc travelled by the mass, the velocity is ds/dt where
s = Lθ(t) is arclength. The acceleration is Lθ ′′ (t). Newton’s second law
for the force along this arc is F = mLθ ′′ (t). Another relation for the
force can be found by resolving the vector gravitational force m~g into
its normal and tangential components. By trigonometry, the tangential
component gives a second force equation F = −mg sin θ(t). Equating
competing forces and cancelling m results in the pendulum equation
(2)
θ ′′ (t) +
g
sin θ(t) = 0.
L
Because the mass m cancels from the equation, the pendulum oscillation
depends only upon the length of the string and not upon the mass!
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θ
m
m~g
Figure 5. A simple pendulum
The approximation sin u ≈ u, valid for small angles u, is applied to (2).
The result is the linearized pendulum
g
θ ′′ (t) + θ(t) = 0.
(3)
L
This equation is indistinguishable from the classical harmonic oscillator, except for variable names. The characteristic solution is θ(t) =
A cos(ωt − α), ω 2 = g/L.
The physical pendulum. The compound pendulum or physical pendulum is a rigid body of total mass m having center of mass C
which is suspended from a fixed origin O – see Figure 6.
O
θ
C
m~g
Figure 6. The physical pendulum
The distance from O to C is denoted d > 0. The gravity vector ~g
makes angle θ with segment OC and it supplies a restoring torque of
magnitude F = −mgd sin θ. Newton’s second law gives a second force
equation F = Iθ ′′ (t) where I is the torque of the rigid body about O.
Force competition results in the compound pendulum
mgd
(4)
sin θ(t) = 0.
θ ′′ (t) +
I
Using sin u ≈ u gives a harmonic oscillator known as the linearized
compound pendulum
(5)
θ ′′ (t) + ω 2 θ(t) = 0,
ω=
s
mgd
.
I
The swinging rod. As depicted in Figure 7, a swinging rod is a
special case of the compound pendulum.
Figure 7. The swinging rod
5.6 Unforced Mechanical Vibrations
219
The torque I = mL2 /3 and center of mass distance d = L/2 are routine
calculus calculations. Then mgd/I = 3mgL/2mL2 = 3g/2L. Applying
(4) gives the swinging rod
θ ′′ (t) +
(6)
3g
sin θ(t) = 0
2L
and applying (5) gives the linearized swinging rod
′′
2
θ (t) + ω θ(t) = 0,
(7)
ω=
r
3g
.
2L
The torsional pendulum. A model for a balance wheel in a watch,
a gavanometer or a Cavendish torsional balance is the torsional pendulum, which is a rigid body suspended by a wire – see Figure 8. The
twisted wire exerts a restoring force F = −κθ0 when the body is rotated
through angle θ0 . There is no small angle restriction on this restoring
force, because it acts in the spirit of Hooke’s law like a linear spring
restoring force. The model uses the Newton’s second law force relation F = Iθ0′′ (t), as in the physical pendulum, but the restoring force is
F = −κθ0 , giving the torsional pendulum
θ0′′ (t)
(8)
θ0
2
+ ω θ0 (t) = 0,
ω=
r
κ
.
I
Figure 8. The torsional pendulum, a
model for a balance wheel in a watch.
The wheel rotates angle θ0 about the
vertical axis, which acts as a spring,
exerting torque I against the rotation.
Shockless auto. An auto loaded with several occupants is supported
by four coil springs, as in Figure 9, but all of the shock absorbers are
worn out. The simplistic model mx′′ (t) + kx(t) = 0 will be applied.
The plan is to estimate the number of seconds it takes for one complete
oscillation. This is the time between two consecutive bottom–outs of the
auto.1
Figure 9. A model for a car on four
springs
Assume the car plus occupants has mass 1350 Kg. Let each coil spring
have Hooke’s constant k = 20000 Newtons per meter. The load is divided
1
Los Angeles patrol officers gave traffic citations to teenagers who disabled the
shocks on their cars. The extreme bronc–buster ride distinguished the modified cars
cruising California boulevards.
220
among the four springs equally, so each spring supports m = 1350/4 Kg.
We will find the natural frequency of vibration ω. Then the number of
seconds for one complete oscillation is the period T = 2π/ω seconds.
The oscillation model for one spring is
1350 ′′
x (t) + 20000x(t) = 0.
4
In the harmonic oscillator form x′′ + ω 2 x = 0, ω 2 =
therefore ω = 7.70, T = 2π/ω = 0.82 seconds.
20000(4)
1350
= 59.26 and
Rolling wheel on a spring. A wheel of total mass m and radius R
is attached at its center to a spring of Hooke’s constant k, as in Figure
10. The wheel rolls without slipping.
k
0
x
Figure 10. A rolling wheel attached to
a spring.
Modelling. Let x(t) be the elongation of the spring from equilibrium,
x > 0 corresponding to the wheel rolling to the right and x < 0 corresponding to the wheel rolling to the left.
If the wheel slides frictionless, then the model is mx′′ (t)+ kx(t) = 0. But
a wheel that rolls without slipping has inertia, and consideration of this
physical difference will be shown to give the rolling wheel equation
(9)
mx′′ (t) +
2
kx(t) = 0.
3
A curious consequence is that x(t) is identical to the frictionless sliding
wheel with spring constant reduced from k to 2k/3. This makes sense
physically, because the action of the rolling wheel is to reduce the stiffness
of the spring.
The derivation is based upon the energy conservation law
Kinetic + Potential = constant.
The kinetic energy T is the sum of two energies, T1 = 12 mv 2 for translation and T2 = 21 Iω 2 for the rolling wheel, whose inertia is I = 12 mR2 .
The velocity is v = Rω = x′ (t). Algebra gives T = T1 + T2 = 34 mv 2 .
The potential energy is K = 12 kx2 for a spring of Hooke’s constant k.
Application of the energy conservation law T + K = c gives the equation
1
3
′
2
2
4 m(x (t)) + 2 k(x(t)) = c. Differentiate this equation on t to obtain
3
′
′′
′
′
2 mx (t)x (t) + kx(t)x (t) = 0, then cancel x (t) to give (9).
5.6 Unforced Mechanical Vibrations
221
Exercises 5.6
Simple harmonic motion. Deter-
7. x′′ + 4x = 0, x(0) = 1, x′ (0) = −1
′′
mine the model equation mx (t) +
kx(t)
= 0, the natural frequency ω =
p
k/m, the period 2π/ω and the solution x(t) for the following spring–mass
systems.
8. x′′ + 4x = 0, x(0) = 1, x′ (0) = 1
9. x′′ + 16x = 0, x(0) = 2, x′ (0) =
−1
1. A mass of 4 Kg attached to 10. x′′ + 16x = 0, x(0) = −2, x′ (0) =
a spring of Hooke’s constant 20
−1
Newtons per meter starts from
′′
′
equilibrium plus 0.05 meters with 11. 5x +11x = 0, x(0) = −4, x (0) =
1
velocity 0.
2. A mass of 2 Kg attached to 12. 5x′′ +11x = 0, x(0) = −4, x′ (0) =
a spring of Hooke’s constant 20
−1
Newtons per meter starts from
equilibrium plus 0.07 meters with 13. x′′ + x = 0, x(0) = 1, x′ (0) = −2
velocity 0.
14. x′′ + x = 0, x(0) = −1, x′ (0) = 2
3. A mass of 2 Kg is attached to a
spring that elongates 20 centimePendulum. The formula
ters due to a force of 10 Newtons.
r
The motion starts at equilibrium
P1
R1 L1
=
with velocity −5 meters per secP2
R2 L2
ond.
is valid for the periods P1 , P2 of two
4. A mass of 4 Kg is attached to a
pendulums of lengths L1 , L2 located
spring that elongates 20 centimeat distances R1 , R2 from the center of
ters due to a force of 12 Newtons.
the earth. The formula implies that a
The motion starts at equilibrium
pendulum can be used to find the rawith velocity −8 meters per secdius of the earth at a location. It is
ond.
also useful for designing a pendulum
5. A mass of 3 Kg is attached to clock adjustment screw.
a coil spring that compresses 2
using
centimeters when 1 Kg rests on 15. Derive the formula,
the
gravitational
relation
the top coil. The motion starts
g = GM/R2 and the period
at equilibrium plus 3 centimeters
formula
p P = 2π/ω, where
with velocity 0.
ω = g/L.
6. A mass of 4 Kg is attached to
a coil spring that compresses 2 16. A pendulum clock taken on a voyage loses 2 minutes a day comcentimeters when 2 Kg rests on
pared to its exact timing at home.
the top coil. The motion starts
Determine the altitude change at
at equilibrium plus 4 centimeters
the destination.
with velocity 0.
Phase–amplitude forms. Solve the 17. A pendulum clock with adjustable
given differential equation and report
the solution in phase–amplitude form
x(t) = A cos(ωt − α) with A > 0 and
0 ≤ α < 2π.
length L loses 3 minutes per day
when L = 30 inches. What
length L adjusts the clock to perfect time?
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18. A pendulum clock with adjustable
length L loses 4 minutes per day
when L = 30 inches. What fineness length F is required for a
1/4–turn of the adjustment screw,
in order to have 1/4–turns of the
screw set the clock to perfect time
plus or minus one second per day?